What is the tangent of 1/((x-7)) ?

The tangent of 1/((x-7)) is y=-1/((a_{0)-7)^2}x+(2a_{0}-7)/((a_{0)-7)^2}

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tangent of 1/((x-7))

tangent of

\frac{1}{( x - 7 )}

y = - \frac{1}{( a _{0} - 7 ) ^{2}} x + \frac{2 a _{0} - 7}{( a _{0} - 7 ) ^{2}}

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

(a_{0}, \frac{1}{a _{0} - 7})

Find the slope of

y = \frac{1}{x - 7} : \frac{d y}{d x} = - \frac{1}{( x - 7 ) ^{2}}

EN : T i tl e G e n er a l Eq u a t i o n Sl o p e A t P o in t 2 Eq : m = - \frac{1}{( a _{0} - 7 ) ^{2}}

Find the line with slope m=

- \frac{1}{( a _{0} - 7 ) ^{2}}

and passing through

(a_{0}, \frac{1}{a _{0} - 7}) : y = - \frac{1}{( a _{0} - 7 ) ^{2}} x + \frac{2 a _{0} - 7}{( a _{0} - 7 ) ^{2}}

y = - \frac{1}{( a _{0} - 7 ) ^{2}} x + \frac{2 a _{0} - 7}{( a _{0} - 7 ) ^{2}}

y^{^{'}} - y = e^{t} cos (t), y (0) = 1

\frac{d}{d x} (\frac{5}{x} - \frac{2}{x ^{3}} + \frac{1}{x ^{4}})

\frac{d}{d x} (sin (2 x) - 2 sin (x))

\int cos (x) cot (x) d x

\frac{d}{d x} (x^{2} + e^{2 x y})

What is the tangent of 1/((x-7)) ?
The tangent of 1/((x-7)) is y=-1/((a_{0)-7)^2}x+(2a_{0}-7)/((a_{0)-7)^2}